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Linear ALgebra question

  1. Jan 30, 2010 #1
    1. The problem statement, all variables and given/known data

    Suppose the linear operator T on P1 over R has the matrix

    3 -2
    1 0

    with respect to A = B = {1-x, x}. Find the matrix of T with respect to A' and B' = {2-x, 1}.




    2. Relevant equations



    3. The attempt at a solution

    I don't understand why there needs to be two different sets A and B if they are they same?
    Need A'A (transition matrix from A to A'), so I write A with respect to A' and get
    [1-x]A' = (1, -1)
    [x]A' = (-1, 2)

    Then A'A =
    1 -1
    -1 2

    Then A'A[T]A = [T]A'
    So
    (1 -1) (3 -2) (2 -2)
    (-1 2) (1 0) = (-1 2) = [T]A'

    But it's wrong?
     
  2. jcsd
  3. Jan 30, 2010 #2

    vela

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    If I deciphered your notation correctly, you have a linear operator [itex]T: P_1 \rightarrow P_1[/itex]. The matrix

    [tex]T_{BA}=\begin{bmatrix}3 & -2\\1 & 0\end{bmatrix}[/tex]

    is T's representation where the domain is represented wrt basis A and the image is represented wrt basis B. The matrix

    [tex]U_{A'A}=\begin{bmatrix}1 & -1 \\ -1 & 2\end{bmatrix}[/tex]

    transforms a representation of a vector from the A basis to the A' basis, and what you calculated was

    [tex]T_{BA}U_{A'A} = \begin{bmatrix}2 & -2 \\ -1 & 2\end{bmatrix}[/tex]

    So you're sort of halfway there. What you need is the matrix [itex]U_{AA'}[/itex] that will transform a vector from the A' basis to the A basis, but what you found earlier was the matrix [itex]U_{A'A}[/itex] which goes in the other direction.

    Let's say [itex]\vec{x}_{A'}[/itex] is the representation of vector [itex]\vec{x}[/itex] wrt the A' basis. When you multiply it by [itex]U_{AA'}[/itex], you get [itex]\vec{x}_A = U_{AA'}\vec{x}_{A'}[/itex], its representation wrt the A basis. Then you can multiply it by [itex]T_{BA}[/itex] to get [itex]T(\vec{x})[/itex] wrt to the B basis. Finally, you need to transform the answer from the B basis to the B' basis. In symbols, you're trying to find

    [tex]T_{B'A'} = U_{B'B}T_{BA}U_{AA'}[/itex]

    You've got most of the pieces. You just need to get a few more and put them all together.
     
    Last edited: Jan 30, 2010
  4. Jan 31, 2010 #3
    I understand the rest of your post.. except for

    " [tex]
    T_{BA}=\begin{bmatrix}3 & -2\\1 & 0\end{bmatrix}
    [/tex]

    is T's representation where the domain is represented wrt basis A and the image is represented wrt basis B."

    How did you get that from the question..?
     
  5. Jan 31, 2010 #4

    vela

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    Isn't that what the first sentence you wrote means? I could be wrong. I'm kind of guessing as to what exactly A and B are as you didn't explain the notation you're using.
     
  6. Jan 31, 2010 #5

    HallsofIvy

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    Here's another way to do it. Write the "new" basis vectors, 2-x and 1, in terms of the "old" basis vectors, 1- x and x. That is, find numbers, a, b, c, d, such that 2- x= a(1-x)+ bx and 1= c(1- x)+ dx. Write those numbers as column matrices, [a b]T and [c d]T and multiply by the given matrix. Write the resulting matrices in terms of the "new" basis vectors. The coefficients will be the columns of the matrix in terms of the new basis vectors.

    The idea is that A'x, where A' is the new matrix and x is a vector in terms of the new basis, is equal to CABx where B is convertion to the old basis, A is the original matrix, and C is convertion back to the new basis.
     
  7. Jan 31, 2010 #6
    So when they say ".. has the matrix with respect to A = B = ..." that means it is a transition matrix from something wrt A to something wrt to B? How do I know its not from from B to A?

    I think I understand the idea of combining multiple transition matrices, it's just the language of the question that confused me.
     
  8. Jan 31, 2010 #7

    vela

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    It could be from B to A. It depends on the notational conventions your class or textbook is using. It doesn't matter in this problem since A=B.
     
  9. Jan 31, 2010 #8
    Okay.
    What happens if I can't find a way to get one of the piece? Like if one of the components of the original cannot be represented with respect to the basis I'm trying to transform into?
     
  10. Jan 31, 2010 #9

    vela

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    If you have a basis A for a vector space, by definition, it spans the vector space, so any vector in that space, including all of the vectors in basis B, can be represented as a linear combination of the A basis vectors.
     
  11. Jan 31, 2010 #10
    Okay thank you I think I get it now.
     
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